13.13 \( \Delta \) ivetar бuvexท่s \( \sigma u v a ́ p \tau \eta \sigma \eta ~ f: ~ \mathbb{R} \rightarrow \mathbb{R} \mu \varepsilon \) \( \mathrm{f}(\mathrm{x}) \neq 0 \gamma \mathrm{l} \alpha \kappa \alpha \dot{\alpha} \theta \varepsilon \mathrm{x} \in \mathbb{R} \). a) \( \mathrm{N} \alpha \) ขлодоүїєєє то ópı: \[ \lim _{x \rightarrow-\infty} \frac{f(1) x^{5}-4 x^{3}+2 x-1}{f(3) x^{2}-5 x+3} \] ß) \( \mathrm{N} \alpha \alpha \pi о \delta \varepsilon i \xi \varepsilon \tau \varepsilon \) ó \( \tau \iota \eta \) \( \varepsilon \xi \) iowoŋः: \[ x f(x)=\left(x^{2}-4\right) e^{x} \]

To approach the limit problem you've presented, we can start by analyzing the degrees of the polynomial in the numerator and denominator. In this case, as \(x\) approaches \(-\infty\), the leading terms of the highest degree will dominate the behavior of the function. So, we would take the limit of \(\frac{f(1)x^5}{f(3)x^2}\) which simplifies to \(\frac{f(1)}{f(3)} x^3\). As \(x\) approaches \(-\infty\), this will tend to \(-\infty\) if \(f(1) \neq 0\). For the second function given, \(x f(x) = (x^2 - 4)e^x\), we can determine \(f(x)\) explicitly. By rearranging the equation, we find that \(f(x) = \frac{(x^2 - 4)e^x}{x}\). A good thing to check next is the behavior of \(f(x)\) as \(x\) approaches \(0\) or \(-\infty\), which can provide insights into the overall shape of the function and its trends across the real line. Don't forget to apply L'Hôpital's rule if you encounter an indeterminate form!